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Determine the normal force of a crate on the floor

  1. Mar 31, 2008 #1
    1. The problem statement, all variables and given/known data
    A boy pulls a 50kg crate across a level floor with an applied force of 200N. If this applied force is at an angle of 30 degrees with the horizontal and the coefficient of kinetic friction is .3, determine;
    a) the normal force exerted on the crate by the floor.

    I understand that the normal force =/= force of gravity in this case due to the applied force being at an angle. When I draw out the free body diagram, I get the following: (see the attachment)

    I now know that I need to subtract the applied force vector that is perpendicular to the crate, but in my freebody diagram, both vectors are pointing up...?
     

    Attached Files:

  2. jcsd
  3. Mar 31, 2008 #2

    rock.freak667

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    Homework Helper

    The resultant vertical force is zero.
    That means that the sum of all the forces up is equal to the sum of all the forces down.
    The forces up are the normal reaction and the component of the applied force. The force acting down is the weight. Can you now find the normal reaction?
     
  4. Mar 31, 2008 #3
    Fnet = Fn +Fa - Fg = 0
    Fn + 200sin30˚ - (50kg*9.8m/s/s) = 0
    Fn + 100N - 490N = 0
    Fn - 390N = 0
    Fn = 390N

    .. and that's the answer. Thanks a lot!
     
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